Let $f$ be a twice differentiable function, and let $f(1)=-7$, $f'(1)=0$, and $f''(1)=-2$. What occurs in the graph of $f$ at the point $(1,-7)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $(1,-7)$ is a minimum point. (Choice B) B $(1,-7)$ is a maximum point. (Choice C) C There's not enough information to tell.
Solution: Since $f'(1)=0$, we know that $x=1$ is a critical point. The second derivative test allows us to analyze what happens in the graph of $f$ at this point according to these three cases: If $f''(1)>0$, the graph of $f$ has a minimum point at $x=1$. If $f''(1)<0$, the graph of $f$ has a maximum point at $x=1$. If $f''(1)=0$, the test is inconclusive. [Why is this so?] We are given that $f''(1)=-2<0$. Therefore, $(1,-7)$ is a maximum point.